# Applying the ratio test

Below is a screen shot of a solution to a problem. The part I don't fathom is after the ratio test is applied to the denominator. How can, noting that an+1, (2n-1) become (2n-1)(2n+1) and not just (2(n+1)-1)=2n+1? Related Calculus and Beyond Homework Help News on Phys.org
Mark44
Mentor
Below is a screen shot of a solution to a problem. The part I don't fathom is after the ratio test is applied to the denominator. How can, noting that an+1, (2n-1) become (2n-1)(2n+1) and not just (2(n+1)-1)=2n+1?

View attachment 83898
In an, the denominator is the product ##1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)##. What will be the next factor in the denominator for an + 1?

BTW, when you post a question here, please don't delete the three parts of the homework template.

• shanepitts
In an, the denominator is the product ##1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)##. What will be the next factor in the denominator for an + 1?

BTW, when you post a question here, please don't delete the three parts of the homework template.

Asked in that manner, the next part is 2n+1. But why not just replace the n with (n+1)? And why would the answer be completely different if this approach to problem is taken?

Mark44
Mentor

Asked in that manner, the next part is 2n+1. But why not just replace the n with (n+1)? And why would the answer be completely different if this approach to problem is taken?
It wouldn't. If you replace n by n + 1 in the expression 2n - 1, what do you get?

• shanepitts
It wouldn't. If you replace n by n + 1 in the expression 2n - 1, what do you get?
You get 2n+1 instead of the (2n-1)(2n+1), no?

Mark44
Mentor
You get 2n+1 instead of the (2n-1)(2n+1), no?
You get (2n + 1). The (2n - 1) factor is the one from an.

• shanepitts
You get (2n + 1). The (2n - 1) factor is the one from an.
Thanks,
I'm sorry but I still don't fathom.

Let's say

αn=2n-1 then αn+1 should equal 2(n+1)-1=2n+1

Why the extra 2n-1 in the αn+1?

Mark44
Mentor
Thanks,
I'm sorry but I still don't fathom.

Let's say

αn=2n-1 then αn+1 should equal 2(n+1)-1=2n+1
This is not an, at least as it's defined in post 1.

$$a_n = \frac{x^n}{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)}$$

Now, what is an + 1?
You need to ask yourself how many factors are in the denominator of an? How many are in the denominator of an + 1?
shanepitts said:
Why the extra 2n-1 in the αn+1?

• shanepitts